Swift Array.map closure issue

Question

I want to improve on a closure I wrote using Swift's Array.map function

I'm basically taking an Array and remapping all of its elements using a closure.

// Here's the array:
var numbersArray = [1, 2, 3, 4, 5]

// Here's the closure - it takes whatever number is passed into it and 
// multiplies it by 2.0
var multiplier = {  (currentNum: Int) -> Double in
    let result = Double(currentNum) * 2.0
    return result
}

// And here's how I'm calling it:
var newArray = numbersArray.map(multiplier)

And this works perfectly.

But what if I want to multiply everything by 2.1? or 3.5? or any value? In other words, what if I want to make the amount I multiply by also be a variable? And have it be passed into the closure as a second argument?

I tried adding it to the argument list like this:

var multiplier = {  (currentNum: Int, factor: Double) -> Double in
    let result = Double(currentNum) * factor
    return result
}

and then changing my call to this:

var newArray = numbersArray.map(multiplier, 3.5)

but I'm getting all sorts of errors (and I tried all sorts of variations on this of course.)

What am I doing wrong?

Connor · Accepted Answer · 2014-07-03 01:21:21Z

14

You can use a higher order function to produce a custom function that you can then use with the array's map function. Like this:

var numbersArray = [1, 2, 3, 4, 5]

func multiplier(factor: Double) -> (Int)->Double
{
    return {  (currentNum: Int) -> Double in
        let result = Double(currentNum) * factor
        return result
    }
}

var newArray = numbersArray.map(multiplier(2.5))

answered Jul 3, 2014 at 1:21

Connor

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4 Comments

sirab333 Over a year ago

That's pretty bad-ass :-) Very clever. Is that pretty much the only way to do this?

Connor Over a year ago

It's the only way I can think of at least. I'm not really a master of functional programming though

sirab333 Over a year ago

Fair enough. Even if its not the only or "best" way to do this, its certainly plenty cool. I'll credit you for right answer. Cheers :-)

rickster Over a year ago

No need to be modest, @connor -- that's good functional programming. :) You'll find some examples much like that in the Swift book, IIRC.

Jack Lawrence · Accepted Answer · 2016-12-28 17:58:41Z

Edit: Note: This language feature was removed in Swift 2.

A swift-er way than connor's answer (but along the same lines), is to use a curried function. From The Swift Programming Language->Language Reference->Declarations->Curried Functions and Methods:

A function declared this way is understood as a function whose return type is another function.

So you can simplify this:

func multiplier(factor: Double) -> (Int)->Double
{
    return {  (currentNum: Int) -> Double in
        let result = Double(currentNum) * factor
        return result
    }
}

to this:

func multiplier(factor: Double)(currentNum: Int) -> Double {
    return Double(currentNum) * factor
}

and use it exactly the same way:

let numbersArray = [1, 2, 3, 4, 5]
let multipliedArray = numbersArray.map(multiplier(3.5))

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